IDZ 12.1 – Option 2. Solutions Ryabushko A.P.

Task No. 1. Prove the convergence of the series and find its sum.

Consider the series $\sum_{n=1}^\infty \frac{2^n}{n!}$. To study its convergence, we use D'Alembert's test:

$$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac{2^{n+1}}{(n+1)!} \cdot \frac{n!}{2^n} = \lim_{n\to\infty} \frac{2}{n+1} = 0.$$

Thus, the series converges. To find its sum, we use the formula for the exponent:

$$\sum_{n=1}^\infty \frac{2^n}{n!} = e^2.$$

Task No. 2. Examine the indicated series with positive terms for convergence.

1. $\sum_{n=1}^\infty \frac{n!}{2^n}$

To study the convergence of this series, we use D’Alembert’s test:

$$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac{(n+1)!}{2^{n+1}} \cdot \frac{2^n}{n!} = \lim_{n\to\infty} \frac{n+1}{2} = \infty,$$

that is, the series diverges.

1. $\sum_{n=1}^\infty \frac{1}{n\ln^2 n}$

To study the convergence of this series, we use the integral criterion:

$$\int_2^\infty \frac{dx}{x\ln^2 x} = \left[-\frac{1}{\ln x}\right]_2^\infty = \frac{1}{\ln 2}

that is, the series converges.

1. $\sum_{n=1}^\infty \frac{1}{n\ln n}$

To study the convergence of this series, we also use the integral test:

$$\int_2^\infty \frac{dx}{x\ln x} = \ln(\ln x)\bigg|_2^\infty = \infty,$$

that is, the series diverges.

1. $\sum_{n=1}^\infty \frac{1}{n^\alpha}$, where $0

To study the convergence of this series, we use the integral criterion:

$$\int_1^\infty \frac{dx}{x^\alpha} = \begin{cases}\frac{1}{\alpha-1}, &\text{=\infty}\alpha > 1\\infty, & \text{s} \alpha \leq 1 \end{cases}$$

Thus, the series converges for $\alpha > 1$ and diverges for $\alpha \leq 1$.

1. $\sum_{n=1}^\infty \frac{1}{n(\ln n)^\alpha}$, where $0

To study the convergence of this series, we also use the integral test:

$$\int_2^\infty \frac{dx}{x(\ln x)^\alpha} = \begin{cases}\frac{1}{\alpha-1}\left(\frac{1}{( \ln 2)^{\alpha-1}} - \lim_{x\to\infty}\frac{1}{(\ln x)^{\alpha-1}}\right), &\text{если } \alpha > 1 \ \infty, & \text{text } \alpha \leq 1 \end{cases}$$

Thus, the series converges for $\alpha > 1$ and diverges for $\alpha \leq 1$.

Task No. 7. Examine alternating series for convergence and absolute convergence.

1. $\sum_{n=1}^\infty \frac{(-1)^n}{n}$

To examine the convergence of this alternating series, we use Leibniz’s test: the sequence $\left|\frac{1}{n}\right|$ monotonically decreases to zero. Also, to study absolute convergence, we will use a comparison with the harmonic series:

$$\sum_{n=1}^\infty \left|\frac{(-1)^n}{n}\right| = \sum_{n=1}^\infty \frac{1}{n}$$

This series diverges because it is a harmonic series. Thus, the original sign-alternating series converges, but does not converge absolutely.

1. $\sum_{n=1}^\infty \frac{(-1)^n}{n^\alpha}$, where $0

To examine the convergence of this alternating series, we use Leibniz’s test: the sequence $\left|\frac{1}{n^\alpha}\right|$ monotonically decreases to zero. To study absolute convergence, we use a comparison with a harmonic series:

$$\sum_{n=1}^\infty \left|\frac{(-1)^n}{n^\alpha}\right| = \sum_{n=1}^\infty \frac{1}{n^\alpha}$$

This series converges for $\alpha > 1$ and diverges for $\alpha \leq 1$. Thus, the original alternating series converges for $\alpha > 1$, and for $\alpha \leq 1$ it diverges, but converges absolutely.

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IDZ 12.1 – Option 2. Solutions Ryabushko A.P. is a digital product that contains solutions to problems included in the second version of Individual homework No. 12.1 on mathematical analysis, written by the author Ryabushko A.P. The solutions are presented in the form of an electronic document with a beautiful HTML design, which makes them easy to use.

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This product may be useful for students and teachers who are studying mathematical analysis and want to test their knowledge in practice. The solutions can be used both for independent work and as a teaching aid.

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